Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.lang.Math.*;
public class A121
{
     public static void main(String args[])throws Exception
     {
         Scanner in=new Scanner(System.in);
         PrintWriter pw = new PrintWriter(System.out);
         
         int n=in.nextInt();
         
         int m=(int)Math.sqrt(n*2);
         int a[]=new int[m+1];
         int i,j,k,l;
         for(i=0;i<=m;i++)
         {
             a[i]=i+1;
             
            }
         for(i=1;i<=m;i++)
         {
             a[i]+=a[i-1];
             
            }
         for(i=0;i<=m;i++)
         {
            if(search(n-a[i],a))
            {
            pw.println("YES");pw.flush();System.exit(0);
             }
            //pw.print(a[i]+",");
            }
         pw.println("NO");
         pw.flush();
            
     }
     public static boolean search(int val,int a[])
     {
         int l=a.length;
         int i,j,k;
         int beg=0,end=l-1,mid=(end+beg)/2;
         while(beg<=end)
         {
             if(a[mid]==val)
             {
                 return true;
                }
             else
             {
                 if(a[mid]>val)
                 {
                     end=mid-1;
                    }
                 else
                 {
                     beg=mid+1;
                    }
                }
             mid=(end+beg)/2;
            }
         return false;
        }
}